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Proof step 6 from proof disjoint-nojunk-newref-subset-01 (previous step)

r1 -- _type ∈ st ∧ st ⊆ st1, (r1 ' + refs1) .last = (r ' + refs0) .last, path∃(str, r, (r ' + refs0) .last, st1), validrefs(str, r ' + refs0, st1[r -- _type] .type, st1), validrefs(str, r1 ' + refs1, st[r1 -- _type] .type, st), newref(r, st), st ⊆ st1, newref((r ' + refs0) .last, st) ⊦ (r ' + refs0) .last = jvmref, r = jvmref, r1 = r, junkref(str, r1, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
(r1 ' + refs1) .last = (r ' + refs0) .last, r1 -- _type ∈ st, path∃(str, r, (r ' + refs0) .last, st1), validrefs(str, r ' + refs0, st1[r -- _type] .type, st1), validrefs(str, r1 ' + refs1, st[r1 -- _type] .type, st), newref(r, st), st ⊆ st1, newref((r ' + refs0) .last, st), r1 ∈ st ⊦ (r ' + refs0) .last = jvmref, r = jvmref, r1 = r, junkref(str, r1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st])]]